Need help with small program

**alokrsx** · 05-24-2009

Hello all,
I've been trying to solve a problem here but I'm not able to get the concept right. Please help me out.
Here is the question and the answer is written below.

Q. Write a computer program to tabulate the total occurances of
each of the alphabets A to Z in a string. Allocate a large
array of 100 or more characters for the input. Use gets function
get a string.

Code:

#include<stdio.h>
#include<conio.h>
#include<string.h>
int main()  {
  int a[100],i, b[100],j;
  printf("Enter the sentence:");
  gets(a,100);
  b[26]={97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123}

for(i=0;i<100;i++){
  for(j=0;j<=26;j++){
    if(a[i]==b[j]){
      if(b[j]==97)
        ac++;
      elseif(b[j]==98)
        bc++;
      elseif(b[j]==99)
        cc++;
      elseif(b[j]==100)
        dc++;
      elseif(b[j]==101)
        ec++;
      elseif(b[j]==102)
        fc++;
      elseif(b[j]==103)
        gc++;
      elseif(b[j]==104)
        hc++;   
  getch();
  return 0;
}

Thanks in advance

**Salem** · 05-24-2009

> ac++;
You don't declare these.

Have you considered an array of counters?

**Adak** · 05-24-2009

Originally Posted by alokrsx

Hello all,
I've been trying to solve a problem here but I'm not able to get the concept right. Please help me out.
Here is the question and the answer is written below.

Q. Write a computer program to tabulate the total occurances of
each of the alphabets A to Z in a string. Allocate a large
array of 100 or more characters for the input. Use gets function
get a string.

Code:

#include<stdio.h>
#include<conio.h>
#include<string.h>
int main()  {
  int a[100],i, b[100],j;
  printf("Enter the sentence:");
  gets(a,100);
  b[26]={97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123}

for(i=0;i<100;i++){
  for(j=0;j<=26;j++){
    if(a[i]==b[j]){
      if(b[j]==97)
        ac++;
      elseif(b[j]==98)
        bc++;
      elseif(b[j]==99)
        cc++;
      elseif(b[j]==100)
        dc++;
      elseif(b[j]==101)
        ec++;
      elseif(b[j]==102)
        fc++;
      elseif(b[j]==103)
        gc++;
      elseif(b[j]==104)
        hc++;   
  getch();
  return 0;
}

Thanks in advance

You're close, but not quite there.

Say you had an array like count[] that was all zero's, initially.

Now for every char in the string. (a for or while or do while loop)

12345]* if the char is a letter A-Z or a-z, change it to all upper case.
12345]* Now we want 'A' to correspond to count[0], and here's the heart of the trick,
12345]* so we *subtract* an 'A' from each letter, and then:

Code:

count[letter]++;

} //end of for,while or do while loop

Now every A letter in the string will increment count[0], and every B will do the same for count[1], and C's will increment count[2], etc.

//now print your count[] array remembering to add 'A' to the index number, as you display the results, and you are printing a char, not a digit, in your printf() line of code.

**alokrsx** · 05-25-2009

Thanks for your reply, Adak.

I do not know how to convert lower case letter to upper case other than using the if statement which will be too long.

Sorry, but I'm still not able to understand

**MK27** · 05-25-2009

You already know about ASCII values. You can just add or subtract from those to change case.

Also, there are toupper() and tolower() macros in ctype.h (*nix).

**alokrsx** · 05-26-2009

Got it.
Thanks a lot for your help everyone.

Code:

#include <stdio.h>
#include <stdlib.h>
#include<string.h>

int main()
{
  int i;
  int nAs = 0;
  int nBs = 0;
  int nCs = 0;
  int nDs = 0;
  int nEs = 0;
  int nFs = 0;
  int nGs = 0;
  int nHs = 0;  
  int nIs = 0;
  int nJs = 0;
  int nKs = 0;
  int nLs = 0;
  int nMs = 0;
  int nNs = 0;
  int nOs = 0;
  int nPs = 0;
  int nQs = 0;  
  int nRs = 0;
  int nSs = 0;
  int nTs = 0;
  int nUs = 0;
  int nVs = 0;
  int nWs = 0;
  int nXs = 0;
  int nYs = 0;
  int nZs = 0;  
  char s1[100];
  printf("Enter the string: ");
  scanf("%s",s1);
  for(i=0; i<strlen(s1); i++)
    {
      if (s1[i] == 'a' || s1[i] == 'A')
        nAs++;
      else if (s1[i] == 'B' || s1[i] == 'b')
        nBs++;
      else if (s1[i] == 'C' || s1[i] == 'c')
        nCs++;
      else if (s1[i] == 'D' || s1[i] == 'd')
        nDs++;
      else if (s1[i] == 'E' || s1[i] == 'e')
        nEs++;
      else if (s1[i] == 'F' || s1[i] == 'f')
        nFs++;
      else if (s1[i] == 'G' || s1[i] == 'g')
        nGs++;
      else if (s1[i] == 'H' || s1[i] == 'h')
        nHs++;
      else if (s1[i] == 'I' || s1[i] == 'i')
        nIs++;  
      else if (s1[i] == 'J' || s1[i] == 'j')
        nJs++;
      else if (s1[i] == 'K' || s1[i] == 'k')
        nKs++;
      else if (s1[i] == 'L' || s1[i] == 'l')
        nLs++;
      else if (s1[i] == 'M' || s1[i] == 'm')
        nMs++;
      else if (s1[i] == 'N' || s1[i] == 'n')
        nNs++;
      else if (s1[i] == 'O' || s1[i] == 'o')
        nOs++;
      else if (s1[i] == 'P' || s1[i] == 'p')
        nPs++;
      else if (s1[i] == 'Q' || s1[i] == 'q')
        nQs++;  
      else if (s1[i] == 'R' || s1[i] == 'r')
        nRs++;
      else if (s1[i] == 'S' || s1[i] == 's')
        nSs++;
      else if (s1[i] == 'T' || s1[i] == 't')
        nTs++;
      else if (s1[i] == 'U' || s1[i] == 'u')
        nUs++;
      else if (s1[i] == 'V' || s1[i] == 'v')
        nVs++;
      else if (s1[i] == 'W' || s1[i] == 'w')
        nWs++;
      else if (s1[i] == 'X' || s1[i] == 'x')
        nXs++;
      else if (s1[i] == 'Y' || s1[i] == 'y')
        nYs++;
      else if (s1[i] == 'Z' || s1[i] == 'z')
        nZs++;     
    }
  printf("\n # of A: ",nAs);
  printf("\n # of B: ",nBs);
  printf("\n # of C: ",nCs);
  printf("\n # of D: ",nDs);
  printf("\n # of E: ",nEs);
  printf("\n # of F: ",nFs);
  printf("\n # of G: ",nGs);
  printf("\n # of H: ",nHs);
  printf("\n # of I: ",nIs);
  printf("\n # of J: ",nJs);
  printf("\n # of K: ",nKs);
  printf("\n # of L: ",nLs);
  printf("\n # of M: ",nMs);
  printf("\n # of N: ",nNs);
  printf("\n # of O: ",nOs);
  printf("\n # of P: ",nPs);
  printf("\n # of Q: ",nQs);
  printf("\n # of R: ",nRs);  
  printf("\n # of S: ",nSs);
  printf("\n # of T: ",nTs);
  printf("\n # of U: ",nUs);
  printf("\n # of V: ",nVs);
  printf("\n # of W: ",nWs);
  printf("\n # of X: ",nXs);
  printf("\n # of Y: ",nYs);
  printf("\n # of Z: ",nZs);
  getch();
  return 0;  
}

**matsp** · 05-26-2009

I guess it's a solution, but it's not very practical or neat. As suggested, use an array for the count, and use tolower or toupper, and then use math to index the array.

--
Mats

**edesign** · 05-26-2009

The loop may look something like this...

Code:

  for(i=0; i<strlen(s1); i++)
    {
        for(j='a'; j< 'z'; j++)
        {
            if(s1[i] == j)
                alphabets[j-'a']++;
        }
    }

prior to this use a loop for upper case to lower case and after this use a loop for printf statement ... that will look better I guess ...

Edesign

**quzah** · 05-26-2009

There's a reason we just didn't give them the source code version of the answer you know. It's not that we couldn't do it. It's that we wanted them to try it on their own first.

Quzah.

**edesign** · 05-28-2009

Originally Posted by quzah

There's a reason we just didn't give them the source code version of the answer you know. It's not that we couldn't do it. It's that we wanted them to try it on their own first.

Quzah.

I am no one to think that you people can't write it ... I have learned everything on C board and I would encourage other people to do so .. the only reason I posted that piece of code is that after getting the 'solution' , it was much likely that alokrsx wouldn't bother to look at suggestions made before ...

perhaps by looking at that small piece of code he can realize how brief and neat it can be ...

Anyways I'll consider your point before posting now on ...

Edesign

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